Skip to content
Related Articles

Related Articles

Improve Article

Maximize XOR subsequence possible by equidistant elements from both ends

  • Difficulty Level : Expert
  • Last Updated : 22 Apr, 2021
Geek Week

Given an array A[] of size N, find the Maximum Xor Subsequence such that both A [ i ] and A [ N – i – 1 ] belong to this subsequence, where i ranges between [0, N – 1].

Examples:

Input: N = 8, A [ ] = {1, 2, 3, 4, 5, 6, 7, 8}
Output: 13
Explanation:
Maximum Xor Subsequence is {1, 3, 4, 5, 6, 8}

Input: N = 5, A [ ] = {3, 2, 6, 5, 4}
Output: 6
Explanation:
Maximum Xor Subsequence is {3, 2, 6, 5, 4}

Approach:
Since, both A[i] and A[N-i-1] should be present in the same subsequence, pair them up and then find maximum xor sum. For each valid index i, calculate (A[i] ^ A[N-i-1]) and store it in new array X. Size of this new array will be N/2.



Naive Solution:
The simplest approach after creating array X[] for this problem is to recursively generate all subsequences of X and find the maximum xor subsequence. 

Following is the recursive definition of maxXorSubseq(X, N, i):

maxXorSubseq ( X, N, i ) = MAX ( X[ i ] ^ maxXorSubseq ( X, N, i + 1 ), maxXorSubseq (X, N, i + 1) )
 

The total possible combinations will be 2N.
Time Complexity: O(2N)

Below is the implementation of the above approach:

C++




// C++ implementation
// of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Returns maximum xor
int maxXorSubseq(vector<int> &x, int n,
                                 int i)
{
    if(i == n)
       return 0;
 
    return max(x[i] ^ maxXorSubseq(x, n,
                                   i + 1),
                      maxXorSubseq(x, n,
                                   i + 1));
}
 
vector<int> compute(vector<int> a, int n)
{
    vector<int> x;
 
    // Calculate a[i]^a[n-i-1]
    for(int i = 0; i < n / 2; i++)
       x.push_back(a[i] ^ a[n - i - 1]);
 
    // If n is odd
    if(n & 1)
       x.push_back(a[n / 2]);
 
    return x;
}
 
// Driver code
int main()
{
    int n = 8;
    vector<int> a = { 1, 2, 3, 4,
                      5, 6, 7, 8 };
     
    // Getting new array x
    vector<int> x = compute(a, n);
     
    int mxXor = maxXorSubseq(x, x.size(), 0);
     
    cout << (mxXor);
    return 0;
}
 
// This code is contributed by mohit kumar 29   
Java



// Java implementation
// of the above approach
import java.util.*;
 
class GFG{
 
// Returns maximum xor
static int maxXorSubseq(List<Integer> x, int n,
                                         int i)
{
    if(i == n)
    return 0;
 
    return Math.max(x.get(i) ^ maxXorSubseq(x, n,
                                            i + 1),
                               maxXorSubseq(x, n,
                                            i + 1));
}
 
static List<Integer> compute(List<Integer> a, int n)
{
    List<Integer> x = new ArrayList<Integer>();
 
    // Calculate a[i]^a[n-i-1]
    for(int i = 0; i < n / 2; i++)
        x.add(a.get(i) ^ a.get(n - i - 1));
 
    // If n is odd
    if((n & 1) == 1)
        x.add(a.get(n / 2));
 
    return x;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 8;
    List<Integer> a = Arrays.asList( 1, 2, 3, 4,
                                     5, 6, 7, 8 );
     
    // Getting new array x
    List<Integer> x = compute(a, n);
     
    int mxXor = maxXorSubseq(x, x.size(), 0);
     
    System.out.println((mxXor));
}
}
 
// This code is contributed by offbeat
Python3



# Python3 implementation
# of the above approach
 
# Returns maximum xor
def maxXorSubseq(x, n, i):
     
    if(i == n):
        return 0
         
    return max(
        x[i]^maxXorSubseq(
        x, n, i + 1), maxXorSubseq(
        x, n, i + 1))
 
def compute(a, n):
     
    x = []
     
    # Calculate a[i]^a[n-i-1]
    for i in range(n//2):    
        x.append(a[i]^a[n-i-1])
     
    # If n is odd
    if(n&1):
        x.append(a[n//2])
         
    return x
 
# Driver code
if __name__ =="__main__":
     
    n = 8
    a = [1, 2, 3, 4, 5, 6, 7, 8]
     
    # Getting new array x
    x = compute(a, n)
     
    mxXor = maxXorSubseq(x, len(x), 0)
     
    print(mxXor)
C#



// C# implementation 
// of the above approach 
using System;
using System.Collections.Generic;
class GFG {
     
    // Returns maximum xor
    static int maxXorSubseq(List<int> x, int n, int i)
    {
        if(i == n)
        return 0;
       
        return Math.Max(x[i] ^ maxXorSubseq(x, n, i + 1), maxXorSubseq(x, n, i + 1));
    }
       
    static List<int> compute(List<int> a, int n)
    {
        List<int> x = new List<int>();
       
        // Calculate a[i]^a[n-i-1]
        for(int i = 0; i < n / 2; i++)
            x.Add(a[i] ^ a[n - i - 1]);
       
        // If n is odd
        if((n & 1) == 1)
            x.Add(a[n / 2]);
       
        return x;
    }
 
  static void Main() {
        int n = 8;
        List<int> a = new List<int>{ 1, 2, 3, 4, 5, 6, 7, 8 };
           
        // Getting new array x
        List<int> x = compute(a, n);
           
        int mxXor = maxXorSubseq(x, x.Count, 0);
           
        Console.WriteLine((mxXor));
  }
}
 
// This code is contributed by divyeshrabadiya07
Javascript



<script>
 
    // Javascript implementation
    // of the above approach
     
    // Returns maximum xor
    function maxXorSubseq(x, n, i)
    {
        if(i == n)
        return 0;
        
        return Math.max(x[i] ^ maxXorSubseq(x, n, i + 1),
        maxXorSubseq(x, n, i + 1));
    }
        
    function compute(a, n)
    {
        let x = [];
        
        // Calculate a[i]^a[n-i-1]
        for(let i = 0; i < parseInt(n / 2, 10); i++)
            x.push(a[i] ^ a[n - i - 1]);
        
        // If n is odd
        if((n & 1) == 1)
            x.push(a[parseInt(n / 2, 10)]);
        
        return x;
    }
     
    let n = 8;
    let a = [ 1, 2, 3, 4, 5, 6, 7, 8 ];
 
    // Getting new array x
    let x = compute(a, n);
 
    let mxXor = maxXorSubseq(x, x.length, 0);
 
    document.write((mxXor));
 
</script>
Output:
13
 
Efficient Approach:
Considering the above implementation, the following is a partial recursion tree for input X = [9, 5, 1]
In the above partial recursion tree, maxXorSubseq({1}) is being solved four times. On drawing the complete recursion tree, it can be observed that there are many subproblems that are solved again and again. So this problem has Overlapping Subproblems and recomputation of the same subproblems can be avoided by using Memoization. 


For memoization, create an array dp [ ] array and store:
dp[i] = max(x[i] ^ maxXorSubseq(x, n, i + 1), maxXorSubseq(x, n, idx + 1)
 
This avoids recalculation for previously calculated indices thus optimizing computational complexity.
Below is the implementation of the above approach:
C++



// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
vector<int> dp;
  
// Returns maximum xor sum
int maxXorSubseq(vector<int> x, int n, int idx)
{
    if (idx == n)
    {
        return 0;
    }
      
    // If already precomputed
    if (dp[idx] != -1)
    {
        return dp[idx];
    }
    int ans = 0;
    ans = max(ans, x[idx] ^ maxXorSubseq(x, n, idx + 1));
    ans = max(ans, maxXorSubseq(x, n, idx + 1));
                     
    // Store the maximum
    dp[idx] = ans;
    return ans;
}
  
vector<int> compute(int a[],int n)
{
    vector<int> x;
      
    // Calculate a[i]^a[n-i-1]
    for(int i = 0; i < n / 2; i++)
    {
        x.push_back(a[i] ^ a[n - i - 1]);
    }
      
    //  If n is odd
    if ((n & 1) != 0)
    {
        x.push_back(a[n / 2]);
    }
    return x;
}
 
// Driver code
int main()
{
    int n = 8;
    int a[] = { 1, 2, 3, 4, 5, 6, 7, 8 };
      
    // Getting new array x
    vector<int> x = compute(a, n);
      
    // Initialize dp array
    for(int i = 0; i < x.size(); i++)
    {
        dp.push_back(-1);
    }
    int mxXor = maxXorSubseq(x, x.size(), 0);    
    cout << mxXor << endl;
    return 0;
}
 
// This code is contributed by divyesh072019.
Java



// Java implementation of the above approach
import java.io.*;
import java.util.*;
 
class GFG{
     
static Vector<Integer> dp = new Vector<Integer>();
 
// Returns maximum xor sum
static int maxXorSubseq(Vector<Integer> x,
                        int n, int idx)
{
    if (idx == n)
    {
        return 0;
    }
     
    // If already precomputed
    if (dp.get(idx) != -1)
    {
        return dp.get(idx);
    }
    int ans = 0;
    ans = Math.max(ans, x.get(idx) ^
                   maxXorSubseq(x, n, idx + 1));
    ans = Math.max(ans,
                   maxXorSubseq(x, n, idx + 1));
                    
    // Store the maximum
    dp.set(idx,ans);
    return ans;
}
 
static Vector<Integer> compute(int[] a,int n)
{
    Vector<Integer> x = new Vector<Integer>();
     
    // Calculate a[i]^a[n-i-1]
    for(int i = 0; i < n / 2; i++)
    {
        x.add(a[i] ^ a[n - i - 1]);
    }
     
    //  If n is odd
    if ((n & 1) != 0)
    {
        x.add(a[n / 2]);
    }
    return x;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 8;
    int[] a = { 1, 2, 3, 4, 5, 6, 7, 8 };
     
    // Getting new array x
    Vector<Integer> x = compute(a, n);
     
    // Initialize dp array
    for(int i = 0; i < x.size(); i++)
    {
        dp.add(-1);
    }
    int mxXor = maxXorSubseq(x, x.size(), 0);
     
    System.out.println(mxXor);
}
}
 
//  This code is contributed by avanitrachhadiya2155
Python3



# Python3 implementation
# of the above approach
 
# Returns maximum xor sum
def maxXorSubseq(x, n, idx):
     
    if(idx == n):
        return 0
         
    # If already precomputed
    if(dp[idx]!=-1):
        return dp[idx]
 
    ans = 0
     
    ans = max(
        ans, x[idx] ^ maxXorSubseq(
                    x, n, idx + 1))
     
    ans = max(ans, maxXorSubseq(
                    x, n, idx + 1))
     
    # Store the maximum
    dp[idx] = ans
     
    return ans
 
def compute(a, n):
     
    x = []
     
    # Calculate a[i]^a[n-i-1]
    for i in range(n//2):
        x.append(a[i]^a[n-i-1])
     
    # If n is odd
    if(n&1):
        x.append(a[n//2])
         
    return x
     
# Declared dp[] array globally
dp = []
 
# Driver code
if __name__ =="__main__":
     
    n = 8
    a =[1, 2, 3, 4, 5, 6, 7, 8]
     
    # Getting new array x
    x = compute(a, n)
 
    # Initialize dp array
    dp = [-1 for i in range(len(x))]
     
    mxXor = maxXorSubseq(x, len(x), 0)
     
    print(mxXor)
    
C#



// C# implementation of the above approach
using System;
using System.Collections.Generic;
class GFG
{
    static List<int> dp = new List<int>();
   
    // Returns maximum xor sum
    static int maxXorSubseq(List<int> x, int n, int idx)
    {
        if (idx == n)
        {
            return 0;
        }
       
        // If already precomputed
        if(dp[idx] != -1)
        {
            return dp[idx];
        }
        int ans = 0;
        ans = Math.Max(ans, x[idx]^maxXorSubseq(x, n, idx + 1));
        ans = Math.Max(ans, maxXorSubseq(x, n, idx + 1));
       
        // Store the maximum
        dp[idx] = ans;
        return ans;
    }
    static List<int> compute(int[] a,int n)
    {
        List<int> x = new List<int>();
       
        // Calculate a[i]^a[n-i-1]
        for(int i = 0; i < n / 2; i++)
        {
            x.Add(a[i] ^ a[n - i - 1]);
        }
       
        //  If n is odd
        if((n & 1) != 0)
        {
            x.Add(a[n / 2]);
        }
        return x;
    }
   
    // Driver code
    static public void Main ()
    {
        int n = 8;
        int[] a = { 1, 2, 3, 4, 5, 6, 7, 8 };
       
        // Getting new array x
        List<int> x = compute(a, n);
       
        // Initialize dp array
        for(int i = 0; i < x.Count; i++)
        {
            dp.Add(-1);
        }
        int mxXor = maxXorSubseq(x, x.Count, 0);
        Console.WriteLine(mxXor);
    }
}
 
// This code is contributed by rag2127
Javascript



<script>
    // Javascript implementation of the above approach
     
    let dp = [];
    
    // Returns maximum xor sum
    function maxXorSubseq(x, n, idx)
    {
        if (idx == n)
        {
            return 0;
        }
        
        // If already precomputed
        if(dp[idx] != -1)
        {
            return dp[idx];
        }
        let ans = 0;
        ans = Math.max(ans, x[idx]^maxXorSubseq(x, n, idx + 1));
        ans = Math.max(ans, maxXorSubseq(x, n, idx + 1));
        
        // Store the maximum
        dp[idx] = ans;
        return ans;
    }
     
    function compute(a, n)
    {
        let x = [];
        
        // Calculate a[i]^a[n-i-1]
        for(let i = 0; i < parseInt(n / 2, 10); i++)
        {
            x.push(a[i] ^ a[n - i - 1]);
        }
        
        //  If n is odd
        if((n & 1) != 0)
        {
            x.push(a[n / 2]);
        }
        return x;
    }
     
    let n = 8;
    let a = [ 1, 2, 3, 4, 5, 6, 7, 8 ];
 
    // Getting new array x
    let x = compute(a, n);
 
    // Initialize dp array
    for(let i = 0; i < x.length; i++)
    {
      dp.push(-1);
    }
    let mxXor = maxXorSubseq(x, x.length, 0);
    document.write(mxXor);
 
 
// This code is contributed by suresh07.
</script>
Output:
13
 
Time complexity: O(N)
 
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.
In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.



My Personal Notes
arrow_drop_up
Next

Practice Questions for Recursion | Set 3
Recommended Articles
Page :
Maximize sum of K pairs made up of elements that are equidistant from both ends of the array
18, Jun 21
Maximize length of longest non-decreasing array possible by the elements present in either ends of the given array
11, Nov 20
Longest subsequence possible that starts and ends with 1 and filled with 0 in the middle
31, May 20
Minimize length of an array by removing similar subarrays from both ends
09, Feb 21
Rearrange the Array to maximize the elements which is smaller than both its adjacent elements
24, Jun 21
Count non-equidistant triplets of distinct array elements having indices in increasing order
12, Jan 21
Minimize replacements by integers up to K to make the sum of equidistant array elements from the end equal
25, Dec 20
Find the sum of all array elements that are equidistant from two consecutive powers of 2
11, May 21
Count pairs with Bitwise XOR greater than both the elements of the pair
06, Nov 20
Maximize count of pairs whose bitwise XOR is even by replacing such pairs with their Bitwise XOR
22, Feb 21
Number of cells in matrix which are equidistant from given two points
08, Jan 20
Total number of triplets (A, B, C) in which the points B and C are Equidistant to A
08, May 20
Bitwise XOR of same indexed array elements after rearranging an array to make XOR of same indexed elements of two arrays equal
05, Aug 21
Maximum flips possible such that no pair of adjacent elements are both 1
19, Jan 21
Maximize length of increasing subsequence possible by replacing array element by nearest primes
04, Oct 20
Maximize length of subsequence consisting of single distinct character possible by K increments in a string
13, May 21
Check if a pair of integers from two ranges exists such that their Bitwise XOR exceeds both the ranges
11, Jun 21
Minimum elements to be removed from the ends to make the array sorted
27, Dec 19
Remove minimum elements from ends of array so that sum decreases by at least K | O(N)
02, Jan 20
Minimum number of array elements from either ends required to be subtracted from X to reduce X to 0
04, Mar 21
Lexicographically largest permutation by sequentially inserting Array elements at ends
15, Oct 21
Maximum length subsequence such that adjacent elements in the subsequence have a common factor
11, Feb 19
Length of the longest subsequence such that xor of adjacent elements is non-decreasing
15, Nov 19
Length of the longest subsequence such that XOR of adjacent elements is equal to K
25, Jan 21
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
ghoshashis545
@ghoshashis545
Vote for difficulty
Current difficulty :
Expert





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox